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2 years ago

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Nathan is saving up to buy a new phone. He already has $90 and can save an additional $10 per week using money from his after sc

hool job. How much total money would Nathan have after 4 weeks of saving? Also, write an expression that represents the amount of money Nathan would have saved in w weeks.

1 answer:

Galina-37 [17]2 years ago

5 0

Answer:

$130

Step-by-step explanation:

If Nathan already has $90 then you would add the amount of money collected after the 4 weeks. In each week he makes $10.

The equation would be $90+($10x4)= $X

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If f(x)= -6x+6 then f^-1(x)=

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Recall that, to get the inverse relation of any expression, we start off by doing a quick switcharoo on the variables, then we solve for "y".

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3 years ago

The Oregon Department of Health web site provides information on the cost-to-charge ratio (the percentage of billed charges that

Alekssandra [29.7K]

Answer:

We conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

Step-by-step explanation:W

We are given with the cost-to-charge ratios for both inpatient and outpatient care in 2002 for a sample of six hospitals in Oregon below;

Hospital 2002 Inpatient Ratio 2002 Outpatient Ratio

1 68 54

2 100 75

3 71 53

4 74 56

5 100 74

6 83 71

Let $\mu_1$ = <u><em>mean cost-to-charge ratio for outpatient care</em></u>

$\mu_2$ = <u><em>mean cost-to-charge ratio for impatient care</em></u>.

SO, Null Hypothesis, $H_0$ : $\mu_1 \geq \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}

Alternate Hypothesis, $H_A$ : $\mu_1 < \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care}

The test statistics that would be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1_+_n_2_-_2$

where, $\bar X_1$ = sample mean cost-to-charge Outpatient Ratio = $\frac{\sum X_1}{n_1}$ = 63.83

$\bar X_2$ = sample mean cost-to-charge Impatient Ratio = $\frac{\sum X_2}{n_2}$ = 82.67

$s_1$ = sample standard deviation for Outpatient Ratio = $\sqrt{\frac{\sum (X_1-\bar X_1 )^{2} }{n_1-1} }$ = 10.53

$s_2$ = sample standard deviation for Impatient Ratio = $\sqrt{\frac{\sum (X_2-\bar X_2 )^{2} }{n_2-1} }$ = 14.33

$n_1$ = sample of hospital for outpatient care = 6

$n_2$ = sample of hospital for outpatient care = 6

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(6-1)\times 10.53^{2}+(6-1)\times 14.33^{2} }{6+6-2} }$ = 12.574

So, <u><em>the test statistics</em></u> = $\frac{(63.83-82.67)-(0)}{12.574 \times \sqrt{\frac{1}{6}+\frac{1}{6} } }$ ~ $t_1_0$

= -2.595

The value of t test statistics is -2.595.

<u>Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.</u>

Since, our test statistics is less than the critical value of t as -2.595 < -1.812, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

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Answer:

Range is [1, ∞)

Step-by-step explanation:

The values of x can be 1 up to infinity (the domain).

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